Question

Solve the following integral:
\[\int{\sqrt{1+\sin 2xdx}}=\]
(a) sin x + cos x + c
(b) sin x – cos x + c
(c) cos x – sin x + c
(d) None of these

Answer 1

Hint: In order to find the solution of this question, we need to remember a few trigonometric identities like \[{{\sin }^{2}}x+{{\cos }^{2}}x=1,\sin 2x=2\sin x\cos x\] and a few more algebraic identities like \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and then a few standard integrations like \[\int{\sin xdx=-\cos x+c}\] and \[\int{\cos xdx=\sin x+c}.\] By using these formulas, we can find the correct answer to this question.

Complete step-by-step answer:
In this question, we have been asked to integrate the function \[\sqrt{1+\sin 2x}\] with respect to x. So, to solve this question, we will first consider \[\int{\sqrt{1+\sin 2x}}dx\] as I. So, we can write,
\[I=\int{\sqrt{1+\sin 2x}}dx\]
Now, we know that, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1.\] So, we get,
\[I=\int{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}dx}\]
Now, we know that 2 sin x cos x = sin 2x. So, we can get,
\[I=\int{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}dx}\]
Now, we know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.\] So, for a = sin x and b = cos x, we get,
\[{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x={{\left( \sin x+\cos x \right)}^{2}}\]
Therefore, we can write I as,
\[I=\int{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}dx}\]
Now, we know that, \[\sqrt{{{a}^{2}}}=a.\] So, we can write I as,
\[I=\int{\left( \sin x+\cos x \right)dx}\]
And we know that it can be further written as,
\[I=\int{\sin xdx}+\int{\cos x}dx\]
We also know that, \[\int{\sin xdx}=-\cos x+{{c}_{1}}\text{ and }\int{\cos xdx}=\sin x+{{c}_{2}}.\] Therefore, we can write I as,
\[I=-\cos x+\sin x+{{c}_{1}}+{{c}_{2}}\]
\[I=\sin x-\cos x+c\]
Hence, we can say, \[\int{\sqrt{1+\sin 2x}dx=\sin x-\cos x+c}.\]
Therefore, we get the option (b) as the right answer.

Note: While solving the question, one can think of differentiating each option one by one and then see which matches the question but it could be complicated as well as a lengthy method and we can also make calculation mistakes by this method. So, it is better to use the conventional method and get the right answer.